Recently, I have been reading Lin and Wang's book titled "The Analysis of Harmonic Maps and their Heat Flows". In the proof of Theorem 2.1.7, by Fubini's theorem, there exists $r_1\in(\frac{r}{2},r)$ such that (see page.12 (2.5)) $$ r_1\int_{\partial B_{r_1}(x)}|\nabla u|^2(r_1,\theta)d\theta\le 8\int_{B_{r}(x)\backslash B_{\frac{r}{2}}(x)}|\nabla u|^2(y)dy. $$ Here $B_r(x)$ is the ball with center $x$ and radius $r$ in $\mathbb{R^2}$ and $u$ is a $W^{1,2}$-vector-valued function.
I tried to use polar coordinates and the mean value theorem of integrals, but I can't see why this estimate holds.
Maybe your confusion comes from the fact that $d\theta$ here should probably refer to the arc length element of the circle of radius $r_1$, not the arc length element of the unit circle. You get an extra factor of $r_1$ otherwise.
Indeed, let $f(r) = \int_{\partial B_r(x)}|\nabla u(y)|^2 dS(y)$, where $dS(y)$ is the arc length element of $\partial B_r(x)$. By the mean value theorem and polar coordinates formula, there exists $r/2 \leq r_1 \leq r$ such that
$$\int_{B_r(x)\setminus B_{r/2}(x)} |\nabla u(y)|^2 dy = \int_{r/2}^r f(r) dr = \frac{r}{2}f(r_1)\geq \frac{r_1}{2} f(r_1)$$
The slightly larger constant in your stated bound might be due to choice of normalization for the arc length measure.